anonymous
  • anonymous
Derivative of x^2:
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
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anonymous
  • anonymous
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ash2326
  • ash2326
y=x^2 dy/dx=2x we know derivative of x^n=nx^(n-1)

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anonymous
  • anonymous
I'm not asking :) hold on
anonymous
  • anonymous
That tangent line has an equation \[ y - x_0^2 = m(x-x_0) \] for some m. Let x > x0. Then, since the points on the line are less than the points on the curve, \[ x^2 \geq m(x-x_0) + x_0^2\] \[x^2-x_0^2 = (x+x_0)(x-x_0) \geq m(x-x_0) \] and x-x0 is positive, so \[x+x_0 \geq m\] Next, for x < x0, the points on the line are likewise less than the points of the curve, so we get the same equation but now x-x0 < 0 so we obtain \[x+x_0 \leq m\] now, for the first one, since x must be greater than or equal to x0, we can say that \[x+x_0 \geq 2x_0 \geq m\] and Likewise for the second, since x must be less than x0, \[x + x_0 \leq 2x_0 \leq m\] thus \[2x_0 \leq m \leq 2x_0\] and thus \[m = 2x_0\] so the slope of the tangent line to the curve y=x^2 at some point x0 is equal to 2x0, no difference quotients or limits required. Ta da!
anonymous
  • anonymous
Don't judge me, I was bored and decided to try a derivative without a limit. Shush. :)
anonymous
  • anonymous
It's really interesting, actually.

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